Calculating Limits That Are Disguised Derivatives

Instructor: Gerald Lemay

Gerald has taught engineering, math and science and has a doctorate in electrical engineering.

The definition of the derivative involves the operation of taking a limit. In this lesson we look at how to evaluate limits in those cases where the limit problem is a derivative in disguise.

Limits As Disguised Derivatives

At the masquerade party you want to ask your friend Deriv for the details of his favorite cooking recipe. But Deriv is disguised as the fictitious comic book character Lim. If you can identify Deriv, you will get the information much quicker.

The definition of the derivative involves taking limits. Sometimes, a limit problem is actually a statement of a derivative. In this lesson we look at how to evaluate limits in those cases where the limit problem is a derivative in disguise. And just like with Deriv, there are various disguises.

f '(x) in Various Forms

A common way to define the derivative f '(x) is with a Δx:


Replacing Δx with h we get:


Instead of h, any letter will do. Like c:


These are three ways to express the derivative using limits.

Disguises for f '(x)

How is a limit evaluated? There are cases where substituting Δx with the limit value gives an answer and we are done. Other times, substitution won't work because it gives zero in the denominator. There are other cases where expanding and simplifying the expression will work but this strategy may be very time consuming if the functions are complicated. For example, evaluating:


The limit value is zero. Substituting zero for Δx gives zero in the denominator. However, expanding the numerator and simplifying, allows the Δx in the denominator to cancel:


This function is not too complicated so this approach works. What if 3(x + Δx)^2 was 3(x + Δx)^5? Simplifying is much more complicated. This lesson is about a better way to find these types of limits by identifying f(x). Then, the limit problem answer is f '(x).

Really? Let's check. If f(x) is 3x^2 then replacing x with x + Δx means f(x + Δx) is 3(x + Δx)^2. The numerator of our example is f(x + Δx) - f(x) which is the first of the three definitions for the derivative.

The beauty of all of this is knowing f(x).

With f(x) identified as 3x^2, the derivative is easy. The derivative of 3x^2 is 6x. Done! The complicated looking limit was a derivative in disguise. The limit evaluates to 6x.

Try evaluating:


Compare this to f(x + h) - f(x). Maybe the f(x) is x^5 - 2x. If this is true, what is f(x + h) ?

Replace the x with x + h. The x^5 - 2x becomes (x + h)^5 - 2(x + h) which is (x + h)^5 - 2x - 2h. See where this is going?

If f(x) is x^5 - 2x, then f(x + h) - f(x) is (x + h)^5 - 2(x + h) - x^5 + 2x which is the numerator of our example.

Okay! f(x) is x^5 - 2x. The derivative of x^5 - 2x is simply 5x^4 - 2. And we are done! Once again, the limit problem was a derivative in disguise. We only have to identify f(x) and take the derivative of f(x) to evaluate the limit. Oh, Oh. Looks like our friend Deriv has more variations of the Lim disguise.

Disguises for f '(a)

Sometimes the derivative of f(x) is evaluated for some value of x. If this value for x is a, then:


This derivate may be disguised:


The disguises for these problems can be quite good. Our friend Deriv is creative. What if f(x) is the square root of x ?

If a is 16, f(a) = f(16) = the square root of 16 = 4. See the 4 in the example?

What about f(x + Δx) ? If f(x) is the square root of x, then f(x + Δx) is the square root of x + Δx.

For x equal to a, f(x + Δx) is the square root of a + Δx.

For a equal to 16, f(a + Δx) is the square root of 16 + Δx. Do you see how f(a + Δx) - f(a) is the numerator of our equation for f '(a) ?

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